A bungee jumper (mass 50kg) is attached to a cord, which is attached to a bridge
ID: 1598467 • Letter: A
Question
A bungee jumper (mass 50kg) is attached to a cord, which is attached to a bridge 100m tall. She falls off the bridge with zero initial velocity. (a) The jumper previously measured the cord to be 10 meters long when un-stretched. What is the speed of the jumper at the point when the cord is at its equilibrium length? (b) As the jumper overshoots the cord's equilibrium length, it acts like a spring, slowing the jumper down. What must the spring constant of the cord be so that the jumper just misses hitting the water? (c) Assume now that it's a windy day, and the jumper feels a lot of air resistance on the way down. Say that instead of making it all the way to the water, the jumper now only makes it 80m down before she stops (assume the same spring constant you found in part b). How much energy has been lost due to air resistance?Explanation / Answer
a) when the rope is unstretched the jumpoer would fall by 10 m
velocity gained is v2 = 2gh
v= sqrt(2*9.8*10) = 14 m/s
b) when the jumber just reaches the water
fall of height h = 100 m
change is PE = mgh = 50 *9.8*100 J
The rope would strecth by 100-10 = 90 m
Pe of the spring = 0.5 kx2 = 0.5 k*902
the jumper would stop when the PE of the rope is = PE of fall
0.5 k*902 = 50 *9.8*100
k = 12.1 N/m
c) when the jumper stopped at 80 m
the rope would stretch by 80-10 = 70 m
KE gained = mgh = 50*9.8*80 = 39200 J
PE of the rope = 0.5*12.1*702 =29645 J
The difference is lot to over come the air resistance
= 39200 -29645 = 9555 J
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